[原]not in/not exists 的 null 陷阱

以前遇到了 not in 子查询的一个 null 陷阱,有经验的朋友可能知道怎么回事了,用代码来说就是:

-- 创建两张测试表:
create table tmp01 as 
with tmp as (
  select 1 as id from dual  union all
  select 2       from dual  union all
  select 3       from dual  union all
  select null    from dual
)
select * from tmp;

create table tmp02 as 
with tmp as (
  select 1 as id from dual  union all
  select 2       from dual  union all
  select null    from dual
)
select * from tmp;

我现在想知道表tmp01有哪些id值不在tmp02中,于是我随手就写了一条语句:

select id
from tmp01 
where id not in ( select id from tmp02 )

我期望的结果是:

         ID
----------
         3

但实际结果却是:

no rows selected

近日读到了dinjun123的大作《符合列NULL问题的研究》,终于静下心来想想这个问题。

 

通常使用 not in / not exists 的场景是希望得到两个集合的“差集”,与真正的差集又略有不同,后文将会提到,一般的写法有两种:

select id from tmp01 where id not in ( select id from tmp02 )
select id from tmp01 where not exists ( select 1 from tmp02 where tmp02.id=tmp01.id )

正如上文提到的例子,第一条语句没有可返回的行(no rows selected),第二条语句返回了结果是:

         ID
----------
    (null)
         3

为什么第一个没有结果呢?

我们可以将第一条语句重写为:

select id from tmp01 where id<>1 and id<>2 and id<>null

id=1或者2的时候很好理解,当id=3的时候,id<>null 的判断结果是UNKNOW,注意不是false,where子句只认true,其他都不认,所以tmp01中没有一个值经过 id<>1 and id<>2 and id<>null 这个长长的条件判断后能获得true,也就不会有结果集返回了。

那第二条语句为什么返回的结果是两条呢?3容易理解,null为什么也在结果集中呢?明明tmp02中有null值的啊,我们仔细看一下子查询的where 子句 tmp02.id=tmp01.id,我们再逐个值来跟踪一下,这里我用笛卡尔乘积来获得结果:

set pagesize 6;
select 
  tmp01.id "tmp01.id" , 
  tmp02.id "tmp02.id" , 
(select case when count(*)>0 
             then '   Yes         ' 
             else '   No          ' 
             end  from dual where tmp01.id=tmp02.id) "Result Exists?" 
from tmp01,tmp02
order by 1,2

结果如下:

  tmp01.id   tmp02.id Result Exists?
---------- ---------- ---------------
         1          1    Yes
         1          2    No
         1      (null)   No

  tmp01.id   tmp02.id Result Exists?
---------- ---------- ---------------
         2          1    No
         2          2    Yes
         2     (null)    No

  tmp01.id   tmp02.id Result Exists?
---------- ---------- ---------------
         3          1    No
         3          2    No
         3     (null)    No

  tmp01.id   tmp02.id Result Exists?
---------- ---------- ---------------
    (null)          1    No
    (null)          2    No
    (null)     (null)    No

从结果来看有这么一个规律:只要 null 参与了比较,Result Exists? 就一定为NO(因为结果是UNKNOW),这个也是关于 null 的基本知识,这就解析了为什么第二条语句的输出是两行。

 

从上面的分析,我们可以“窥视”出 in/not in 的结果是依赖于“=”等值判断的结果;exists/not exists 虽然是判断集合是否为空,但通常里面的子查询做的是值判断。

知道了造成结果集出乎意料的原因,我们就可以修改我们的SQL了,为了测试方便,将原来的表tmp01和tmp02改名:

rename tmp01 to tmp01_with_null;
rename tmp02 to tmp02_with_null;

我们看看测试用例:

 test case id   tmp01 has null   tmp01 has null  result has null
------------- ---------------- ---------------- ----------------
            1             true             true            false
            2             true            false             true
            3            false             true            false
            4            false            false            false

其中test case 4 就是打酱油的,只要SQL没有写错,一般不会出问题。

最终,SQL语句改写为:

-- not in 求差集
with tmp01 as (
  select id from tmp01_with_null --where id is not null
),
tmp02 as (
  select id from tmp02_with_null --where id is not null 
)
-- start here
select id from tmp01 
where id not in ( select id from tmp02 where id is not null )
-- 以下是新加的,应付 test case 2
union all
select null from dual 
where exists ( select 1 from tmp01 where id is null )
and not exists ( select 1 from tmp02 where id is null )

-- not exists 求差集
with tmp01 as (
  select id from tmp01_with_null --where id is not null
),
tmp02 as (
  select id from tmp02_with_null --where id is not null 
)
-- start here
select id from tmp01 
where not exists ( 
  select 1 from tmp02 
  where (tmp02.id=tmp01.id) 
  -- 这行是新加的,应付 test case 1
  or (tmp02.id is null and tmp01.id is null )  
)

写了这么多,有人会提议使用minus操作符:

with tmp01 as (
  select id from tmp01_with_null --where id is not null
),
tmp02 as (
  select id from tmp02_with_null --where id is not null 
)
-- start here 
select id from tmp01
minus
select id from tmp02 

貌似语句很简单,但是结果确不一样,请看下面这条语句:

with tmp01 as (
  select id from tmp01_with_null --where id is not null
  union all                      -- 注意这里,现在tmp01已经有重复行了
  select id from tmp01_with_null -- 注意这里,现在tmp01已经有重复行了
),
tmp02 as (
  select id from tmp02_with_null --where id is not null 
)
-- start here 
select 'minus ' as sql_op,id from tmp01
minus
select 'minus ',id from tmp02 
union all
-- not in
select 'not in',id from tmp01 
where id not in ( select id from tmp02 where id is not null )
union all
select 'not in',null from dual 
where exists ( select 1 from tmp01 where id is null )
and not exists ( select 1 from tmp02 where id is null )
union all
-- not exists
select 'not exists',id from tmp01 
where not exists ( 
  select 1 from tmp02 
  where (tmp02.id=tmp01.id) 
  -- 这行是新加的,应付 test case 1
  or (tmp02.id is null and tmp01.id is null )  
);
SQL_OP             ID
---------- ----------
minus               3
not in              3
not in              3
not exists          3
not exists          3

minus消灭了重复行!这就是前文所说的 not in 和 not exists 并非真正意义上的差集。

刚在博问中发现有位朋友遇到了这个陷阱 一个sql 语句in not in 的问题,不知道大家见到过吗?

posted @ 2010-09-04 17:03  killkill  阅读(...)  评论(...编辑  收藏